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A problem that I have been thinking: For what kind of functions $f:[a,b]\to \mathbb R$ is $x\mapsto\sup_{t\in [a,x]}f(t)$ continuous? This is not true for bounded functions, take for example $f(x)=0$ in $[0,1]$ and $f(x)=1$ in $(1,2]$. It is however trivially true for continuous monotone functions. Assuming mere continuity for $f$ I can show right continuity of $\max_{t\in[a,x]}f(t)$:

For $\epsilon$ and some $\delta>0$, $$x\in [x_0-\delta,x_0+\delta]\implies f(x_0)-\epsilon<f(x)<f(x_0)+\epsilon$$ If $x\in [x_0,x_0+\delta]$ then $$\max_{t\in [a,x]}f(t)=\max\left\{\max_{t\in [a,x_0]}f(t),\max_{t\in [x_0,x]}f(t)\right\}$$ If $max_{t\in [a,x_0]}f(t)\ge \max_{t\in [x_0,x]}f(t)$ then $$\left|\max_{t\in [a,x]}f(t)-\max_{t\in [a,x_0]}f(t)\right|=0< \epsilon$$ Otherwise, $$\left|\max_{t\in [a,x]}f(t)-\max_{t\in [a,x_0]}f(t)\right|\le \max_{t\in [x_0,x]}f(t)-\max_{t\in [a,x_0]}f(t)<f(x_0)+\epsilon-\max_{t\in [a,x_0]}f(t)\le \epsilon$$ which proves right continuity. When I tried to prove left continuity I was stuck on showing $$\max_{t\in [x,x_0]}f(t)-\max_{t\in [a,x]}f(t)<\epsilon$$ Can this be circumvented?

Stefan Hamcke
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Optional
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    If $f$ is continuous the max function is continuous. Your proof of right-continuity is wrong, one cannot replace the max on $[a,x]$ by the max on $[x_0,x]$ like you do. – Did Feb 12 '14 at 21:12
  • @Did The first max is the max of the 2nd and 3rd. If the 2nd is bigger than the 3rd then the result is $0$ which is smaller than $\epsilon$ – Optional Feb 12 '14 at 21:17
  • "If the 2nd is bigger than the 3rd then" you are asserting that the absolute value on the LHS is less than a negative real number. – Did Feb 12 '14 at 21:21
  • @Did I do not see what you mean. If the max over $[a,x_0]$ is bigger than the one on $[x_0,x]$ then the max over $[a,x]$ is the one over $[a,x_0]$ and the absolute value is $0$. Otherwise, you have what I have written under 'hence' – Optional Feb 12 '14 at 21:23
  • A similar question was asked today http://math.stackexchange.com/questions/673540/showing-continuity-of-a-function-that-depends-on-another-continous-function – dani_s Feb 12 '14 at 21:27
  • Yes, then the absolute value is zero and you assert that $0\leqslant\max_{[x_0,x]}-\max_{[a,x_0]}$, possibly negative. – Did Feb 12 '14 at 21:28
  • @Did I assert that in the second case (where the 3rd is bigger than the second). I edited the post to make my point clear – Optional Feb 12 '14 at 21:30

3 Answers3

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I'm going to call your max function "$g(x)$" so I don't have to keep writing it down. I think you've got two cases (which probably can be combined but it's not how the idea popped into my head). If $g(x_0)\neq f(x_0)$, then $g(x_0)-f(x_0)=\eta>0$, and you can use the continuity of $f$ to find a $\delta$ corresponding to $\epsilon=\min(\eta/2,\epsilon_{\text{given}})$ where $f(x)<g(x_0)$ on the interval, and therefore $g(x)$ must be constant on that interval (and therefore continuous). If they are equal, then you can use the $\delta$ corresponding to $\epsilon$ for $f$ to show that $|f(x)-f(x_0)|<\epsilon$ for that $\delta$-interval in $x$. Then since $g(x)$ is non-decreasing and $g(x)\geq f(x)$ for all $x$, you have $f(x)\leq g(x)\leq g(x_0)$ for $x\in(x_0-\delta,x_0)$.

tabstop
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  • Why must it be constant on $[x_0-\delta,x_0+\delta]$. I agree it must be constant on $[x_0,x_0+\delta]$. Isn't this the same problem I am facing with left continuity? – Optional Feb 12 '14 at 21:27
  • Suppose there is a point $x_1$ in the interval with $x_1<x_0$ and $g(x_1)<g(x_0)$ (strictly). Somewhere on $[x_1,x_0]$ we must have $f(x)=g(x_0)$ for that number to be the max when we get to $x_0$; but by construction $f(x) < g(x_0)-\epsilon$ everywhere on the interval. So the value $g(x_0)$ cannot be attained anywhere on the interval, and so that maximum value must have been set before we got to the interval. – tabstop Feb 12 '14 at 21:31
  • I see. This essentially solves the left continuity problem as well. – Optional Feb 12 '14 at 21:38
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Suppose $g:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is continuous and suppose $C$ is compact. Then the function $h(x) = \max_{t \in C} g(x,t)$ is continuous.

Suppose $x_n \to \hat{x}$, and suppose $t_n \in C$ is such that $h(x_n) = g(x_n,t_n) \ge g(x_n,t)$ for all $t \in C$.

Let $t^*$ be any accumulation point of the sequence $t_n$, then by continuity we see that (i) $g(x_{n_k}, t^*_{n_k}) \to g(\hat{x}, t^*)$ and (ii), since $ g(x_n,t_n) \ge g(x_n,t)$ for all $t \in C$, we have $g(\hat{x},t^*) \ge g(\hat{x},t)$ for all $t \in C$, and so $g(\hat{x},t^*) = h(\hat{x})$.

Now consider any subsequence of $\eta_n=h(x_n)$, say $\eta_{n_k}$. Since $C$ is compact, the $t_{n_k}$ have a convergent sub-subsequence, say $t_{n_{k_j}}$ that converges to some $t^*$. As above, we see that $\eta_{n_{k_j}} \to h(\hat{x})$. If follows that the entire sequence converges to $h(\hat{x})$ and so $h$ is continuous.

In terms of the original problem, let $C=[0,1]$ and let $g(x,t) = f(xt)$, which is continuous, hence the $\max$ function is continuous.

copper.hat
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For a more general result see my answer here

To adjust this to your problem, replace $[a,b]$ by $[0,1]$, and set $f(x,t)=f(t):[0,1]\times[0,1]\to\Bbb R$, and $c(x)=x$.

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Stefan Hamcke
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